# How much of each substance do I need if I have "100.0 g" of iron?

## My chemistry book says: "Aluminium powder and iron (III) oxide react with each other in a "thermite" reaction, which releases a lot of heat. This reaction has been used (for example) in building railways, which requires molten iron. The chemical equation regarding this reaction is 2 Al(s) + Fe2O3(s) --> 2 Fe(l) + Al2O3(s). How much of each substance is needed, if you have 100,0 grams of iron?" I'm sorry for possible mistakes in my grammar, english is not my first language.

Aug 22, 2017

Here's what I got.

#### Explanation:

You know that the balanced chemical equation that describes this reaction looks like this

$2 {\text{Al"_ ((s)) + "Fe"_ 2"O"_ (3(s)) -> 2"Fe"_ ((l)) + "Al"_ 2"O}}_{3 \left(s\right)}$

Take a look at the coefficients added in front of the four chemical species that are involved in the reaction--keep in mind that if no coefficient is visible, that means that the coefficient is actually equal to $1$.

In this case, you have a $2$ in front of aluminium and a $1$ in front of iron(III) oxide on the reactants' side and a $2$ in front of iron and a $1$ in front of aluminium oxide on the products' side.

This tells you that the reaction consumes $2$ moles of aluminium and $1$ mole of iron(III) oxide and produces $2$ moles of iron and $1$ mole of aluminium oxide.

The problem tells you that the reaction produced $\text{100.0 g}$ of iron, so start by converting the mass of iron to moles. To do that, use the molar mass of iron

100.0 color(red)(cancel(color(black)("g"))) * "1 mole Fe"/(55.845color(red)(cancel(color(black)("g")))) = "1.791 moles Fe"

Now, use the aforementioned mole ratios to find the number of moles of aluminium and of iron(III) oxide needed to produce that many moles of iron

1.791 color(red)(cancel(color(black)("moles Fe"))) * ("1 mole Fe"_2"O"_3)/(2color(red)(cancel(color(black)("moles Fe")))) = "0.8955 moles Fe"_2"O"_3

1.791 color(red)(cancel(color(black)("moles Fe"))) * "2 moles Al"/(2color(red)(cancel(color(black)("moles Fe")))) = "1.791 moles Al"

To convert the number of moles to grams, use the molar masses of the two reactants

$0.8955 \textcolor{red}{\cancel{\textcolor{b l a c k}{{\text{moles Fe"_2"O"_3))) * "159.69 g"/(1color(red)(cancel(color(black)("mole Fe"_2"O"_3)))) = color(darkgreen)(ul(color(black)("143.0 g Fe"_2"O}}_{3}}}}$

$1.791 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{moles Al"))) * "26.98 g"/(1color(red)(cancel(color(black)("mole Al")))) = color(darkgreen)(ul(color(black)("48.32 g}}}}$

The answers are rounded to four sig figs, the number of sig figs you have for the mass of iron.